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Atomic, Planetary, and Biological Systems Described Under A Common Idea
By!
Ian Beardsley!
Copyright © 2024"
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Important To Point Out
You can speak of the structure of the solar system even though it changes with time. This is
important to understand when I refer to sizes of the Moon and the planets, and their orbital
distances.
The whole object of developing a theory for the way planetary systems form is that they meet the
following criterion: They predict the Titius-Bode rule for the distribution of the planets; the
distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The
distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from
the parent star, the outward pressure gradient from the stellar production of radiation, and the
outward inertial forces as a cloud collapses into a flat disc around the central star. These forces
separate the flat disc into rings, agglomerations of material, each ring from which a different
planet forms at its central distance from the star (it has a thickness). In a theory of planetary
formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of
planets today, which was the distribution of the rings from which the planets formed.
of 3 53
Abstract
This paper describes planetary systems, atomic systems, and biological systems under one
idea, as such it is kind of a theory of everything.!
In Part 1 of this paper I show that the system of measuring time we have today, that divides
the day into 24 hours and the hours into 60 minutes, and the minutes into 60 seconds, that
came to us from the ancient Sumerians, who divided the day into 24 hours, and who gave us
the base 60 counting that resulted in our base unit of time, the second, to have the duration it
has today, was a choice that happens to coincide with my theory that shows that our Earth/
Moon/Sun system and atomic systems have a solution in common of the Schrödinger wave
equation with the same base unit of time of 1 second.!
The ancient Sumerians were the first to settle down from wandering, following the herds and
hunting them with stone spearpoint to invent agriculture, architecture, mathematics,
astronomy, and writing. They gave us civilization as we know it today. Since their second came
to us from their system of counting and I have found the second to be a base unit and natural
constant for our planetary system and atomic systems in common, I think it is reasonable to
suggest that perhaps they were given their system of mathematics by ancient Aliens. This is
easy to suggest because their writings say they were given their knowledge by “Those who
came from above.”!
Part 2 I have a theory for the star systems, the planets and their suns, wherein such systems are
solved with the Schrödinger wave equation that is used to solve atomic systems. The result is
that star systems and atomic systems, systems on the macroscales and microscales, are
governed by the same underlying principles. It came to pass that this theory indicated an
overlap with biological systems indicating that biological life could be part of the same idea.
Here I develop that.
Part 3 Our solution of the wave equation for the planets gives the kinetic energy of the Earth
from the mass of the Moon orbiting the Earth, but you could formulate based on the Earth
orbiting the Sun. This yields one second as a base unit as well.
Part 4 We want to find what the wave equation solutions are for Jupiter and Saturn because
they significantly carry the majority of the mass of the solar system and thus should embody
most clearly the dynamics of the wave solution to the Solar System.
Part 5 We want to solve a habitable planetary system in general and apply it to another star
system. We must include the solution of its moon as well, because the Moon of the Earth makes
life optimally possible.
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Contents
Part 1: The Sumerians Give Us The Unit of a Second
Which Turns Out To Be A Natural Constant………………………………………5
Part 2: Biological Life Part of a Universal Natural
Process In The Universe……………………………………………………………….20
Part 3: The Solar Formulation For The Solution
Of The Wave Equation………………………………………………………………….30
Part 4: Wave Solutions For Jupiter And Saturn39
Part 5: Modeling Habitable Star Systems.44
Appendix 1………………………………………………………………………………….50
Appendix 2: The Data For Verifying The Equations………………………………52!
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Part 1: The Sumerians Give Us The Unit of a Second Which Turns Out To Be A Natural
Constant!
of 6 53
While we can measure time by the completion of the Earth in its orbit around the Sun, 365.25
days, or by the rotation of the Earth once in 24 hours giving us the 24 hour day, we measure
the day by a division of it into 24 equal units we call the hour because the ancient Sumerians
divided the day from sunrise to sunset into 12 hours because they could measure time in the
day by the Sun’s position in the sky as determined by a shadow cast by it on the ground that
moves through 12 divisions in a day, each division called an hour. 12 hours are then added on
for the night from sunset to sunrise, so we have the 24 hour day from sunrise to sunrise, or
sunset to sunset. But why choose 12 divisions which gives us the duration of the hour? I
believe the reason is because 12 is the first abundant number. An abundant number means it is
divisible evenly by so many integers preceding it, that their sum is greater than the number
itself. That is,12 is evenly divisible by 1,2,3,4,6 which is greater than 12:!
1+2+3+4+6=16>12!
There is more reason to choose 12 and that is what it is divisible by. Firstly it is divisible by the
two smallest primes 2 and 3, the factors by which no integer can by reduced to further. 2 times
3 is 6 and 12 is evenly divisible by 6 and 6 is of primary importance in mathematics in that the
six-sided regular polygon, the regular hexagon, has its radii equal in lengths to it sides,
something, for one thing, that makes it easy as a tool for computing pi ( ) the ratio of the
circumference of a circle to its diameter. This because inscribed in a circle it approximates the
circle, meaning its perimeter of six, if each side is taken as one, gives its diameter is two,
meaning is close to six divided by two is three. Further twelve is evenly divisible by 4, and the
four-sided square is one of the regular polygons that tessellate meaning the square can tile a
surface without leaving gaps. This makes it suitable for a coordinate system to specify a point
in the plane or space. Such a coordinate system is the first we learn of in school, and is the
most used. iIn fact, other coordinate systems are defined in terms of it like spherical systems
where the angles are measured from a set of rectangular axes.!
The other two regular polygons that tesselate are the equilateral triangle and our regular
hexagon.!
The Sumerians use a base 60 (sexagesimal) counting system because 60 is evenly divisible by
so much as well:!
1,2,3,4,5,6,10, 12, 15, 20, 30!
Making it an abundant number as well, it adds up to 108, which is much greater than 60. It is
important to have 5 because in the regular pentagon it determines the golden ratio, phi ( ),
which is recurrent throughout Nature. Life more often meets up with 5-fold symmetry, like a
starfish with five arms, a human with two legs, two arms, and a head, or five fingers on each
hand. The physical more often meets up with six-fold symmetry, like in the points on a
snowflake. We then have our base unit of one second comes from dividing our day into 24
hours and each of those hours into 60 minutes, and each of those minutes into 60 seconds.
Thus the unit of a second comes to us from the ancient Sumerians, but I find the unit of a
second is a natural constant for the the atom and the solar system. !
So we have to ask did ancient aliens give the Sumerians their method of measuring time with
the 24 hour day, and base 60 sexagesimal counting. Indeed, they were the first to settle down
from hunting with stone spearpoint and following the herds to invent agriculture, architecture,
mathematics, and writing and their myths say they were given this knowledge from “Those who
came from the above”. !
π
π
ϕ
of 7 53
This second I found is the base unit of the orbital dynamics of the solar system. I found I could
create three operators, one that acts on the radius of the proton to its mass, and the other that
acts on the radius of the Sun to its mass, and one that acts on angular momentum to speed.
is Planck’s constant, is the universal constant of gravitation, is the speed of light, is the
fine structure constant, is the radius of a proton, is the mass of a proton, is the mass
of the Moon, is its radius, and the EarthDay is the rotation period of the Earth:!
1)
2)
3)
They seem to suggest that the Earth orbit might by quantized in terms of the Moon and a base
unit of one second.
We say there is a Planck constant for the solar system. We do it with a base unit of one second
because equations 1 and 2 suggest we should . We suggest it is such that it is given by the
standard Planck constant for the atom, , times some constant, , and the Kinetic energy of the
Earth.
4)
5)
Where
6).
Where equation 6 comes from equation 3.
We derive the value of our solar Planck constant
=
h
G
α
r
p
m
p
M
m
R
m
(
1
6α
2
4πh
Gc
)
r
p
m
p
= 1secon d
(
M
m
R
m
(Ear th Da y)
)
R
M
= 1secon d
2
3
π r
p
α
4
G m
3
p
1
3
h
p
c
= 1secon d
h
C
= (hC )K E
e
hC = 1secon d
C =
1
3
1
α
2
c
2
3
π r
p
G m
3
p
C =
1
3
1
α
2
c
1
3
2π r
p
G m
3
p
1
3
18769
299792458
1
3
2π (0.833E 15)
(6.67408E 11)(1.67262E 27)
3
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=
=
=
Now we show that our Planck constant for the solar system gives the base unit of one second for
the quantization. We call our solar Planck constant and find the wavelength for the Moon
which is the ground state for the solar system:
7)
Then wavelength associated with the Moon divided by the speed of light should be 1 second if
our planetary system is quantized in terms of the Moon and a base unit of one second. We have
8)
And we see it is, so we have
9)
The solution for the orbit of the Earth around Sun with the Schrödinger wave equation can be
inferred from the solution for an electron around a proton in the a hydrogen atom with the
Schrödinger wave equation. The Schrödinger wave equation is, in spherical coordinates
1.55976565E 33
s
m
m
kg
3
s
2
kg
m
3
=
s
m
s
2
kg
2
m
2
=
s
m
s
kg m
=
1
kg
s
2
m
2
1
C
= kg
m
2
s
2
=
1
2
mv
2
= energ y
hC = (6.62607E 34)(1.55976565E 33) = 1.03351secon ds 1.0secon ds
hC =
(
kg
m
s
2
m s
)
(
1
kg
s
2
m
2
)
(
kg
m
2
s
)(
1
kg
s
2
m
2
)
= second s
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
= (hC )K E
earth
= (1.03351s)(2.7396E 33J ) = 2.8314E 33J s
λ
moon
=
2
GM
3
m
=
(2.8314E 33)
2
(6.67408E 11)(7.34763E 22kg)
3
= 3.0281E8m
λ
moon
c
=
3.0281E8m
299,792,458m /s
= 1.010secon d s
λ
moon
c
= 1secon d
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10)
Its solution for the atom is
11)
12)
is the energy for an electron orbiting protons and , is the orbital shell for an electron with
protons, the orbital number. I find the solution for the Earth around the Sun utilizes the
Moon around the Earth. This is different than with the atom because planets and moons are not
all the same size and mass like electrons and protons are, and they don’t jump from orbit to
orbit like electrons do. I find that for the Earth around the Sun
13)
14)
is the kinetic energy of the Earth, and is the planet’s orbit. is the radius of the Sun,
is the radius of the Moon’s orbit, is the mass of the Earth, is the mass of the Moon, is
the orbit number of the Earth which is 3 and is the Planck constant for the solar system.
Instead of having protons, we have the radius of the Sun normalized by the radius of
the Moon. It is has always been an amazing fact the the sizes of the Moon and the Sun are such
that given their orbital distances, the Moon as seen from the Earth perfectly eclipses the Sun.
This becomes part of the theory and we suggest it is a condition for sophisticated planets that
harbor life by writing it:
15)
That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is
about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say
the system is quantized by the Moon and the base unit of one second. It is a fact that the Moon
orbiting the Earth optimizes the conditions for life on Earth because it holds the Earth at its
inclination to its orbit around the Sun allowing for the seasons, preventing extreme hot and
extreme cold. Let us now see if our solar Planck constant works…
16)
2
2m
[
1
r
2
r
(
r
2
r
)
+
1
r
2
sinθ
θ
(
sinθ
θ
)
+
1
r
2
sin
2
θ
2
ϕ
2
]
ψ + V(r)ψ = E ψ
E =
Z
2
(k
e
e
2
)
2
m
e
2
2
n
2
r
n
=
n
2
2
Z k
e
e
2
m
e
E
Z
r
n
Z
n
K E
e
= n
R
R
m
G
2
M
2
e
M
3
m
2
2
r
n
=
2
2
GM
3
m
R
R
m
1
n
K E
e
r
n
R
r
m
M
e
M
m
n
Z
R
/R
m
r
planet
r
moon
R
star
R
moon
R
R
m
=
6.96E8m
1737400m
= 400.5986
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=
=2.727E36J
The kinetic energy of the Earth is
The accuracy of our equation is:
Which is very good. Thus we have solved the Earth/Moon/Sun System with our spacetime
operators by using them to find a Planck constant for the solar system.
We want to suggest that the base 60 dynamic applied to the rotational velocity of the Earth to
measure time has a natural property associated with the atom and the Earth/Moon/Sun orbital
system as a solution to the atom’s wave equation. The Earth day is given by, in seconds
This gives since
Thus we have as factors for the seconds per day the smallest primes 2 and 3, and the 4 of
rectangular coordinate systems, and the versatile, abundant, 60. We then suggest the second is
dynamic in terms of what the ancients gave us because 86400 seconds comes from
and this I suggest is connected to Nature because the Earth rotates at a speed
that gives from these ancient factors the duration of a second that I found is the base unit of the
atomic systems and our planetary system, in particular of our Earth/Moon/Sun system as a
solution of the Schrödinger wave equation that solves the same atomic system, which I will go
into right now.
We see the spacetime operators solve the atom by giving us the radius of a proton. We set
equation 3 equal to equation 1
3)
1).
These two yield
K E
e
= (1.732)(400.5986)
(6.67408E 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.727E 36J
2.7396E 33J
100 = 99.5 %
(
1d ay
24hrs
)(
1hr
60min
)(
1min
60sec
)
=
1
86400
d ay
sec
2 3 4 = 24
2 3 4 60 60 = 86400second s /d a y
2 3 4 60 60
2
3
π r
p
α
4
G m
3
p
1
3
h
c
= 1secon d
(
1
6α
2
4πh
Gc
)
r
p
m
p
= 1secon d
of 11 53
17).
I find this is close to the experimental value of the radius of a proton. I find I can arrive at this
radius of a proton another way, energy is given by Plancks constant and frequency
We have
We take the rest energy of the mass of a proton :
The frequency of a proton is
Since our theory gave us the factor of 2/3 for the radius of a proton we have:
The radius of a proton is then
This is close to the CODATA value for the proton radius around 2018 (0.842E-15m). The result
is
We find it may be that the radius of a proton is actually
18)
r
p
=
2
3
h
cm
p
E = h f
f = 1/s, h = J s, h f = (J s)(1/s) = J
E = J = J oules = energ y
m
p
E = m
p
c
2
f
p
=
m
p
c
2
h
m
p
c
2
h
r
p
c
=
2
3
ϕ =
m
p
c
h
r
p
m
p
r
p
=
2
3
h
c
r
p
=
2
3
h
cm
p
r
p
=
2
3
6.62607E 34
(299,792,458)(1.67262E 27)
= 0.88094E 15m
r
p
= ϕ
h
cm
p
= 0.816632E 15m
of 12 53
Where is the golden ratio constant (0.618). This is more along the lines of more recent
measurements. Both equations 17 and 18 for the radius of the proton can be right depending on
the dynamics of what is going on; the radius of a proton is not precisely defined, it is more of a
fuzzy cloud of subatomic particles. Thus we have solved the atom with our spacetime operators
by producing the radius of a proton.
Indeed if the dynamics of the factors the ancients gave us to create the duration of a second are
connected to the dynamics of stars systems then then the second should define the rotational
angular momentum of the Earth since we divide the rotation period of the Earth into these
factors to get the unit of second, and should be connected to our Planck constant for the solar
system which is in units of angular momentum as well, as is Plancks constant for the atom.
The angular momentum of the Earth with respect to the Sun is 2.66E40 kg m2/s. The rotational
angular momentum is 7.05E33 kg m2/s. In orbit angular momentum is given by
For a uniform rotational sphere it is given by
We found our solar system Planck constant was
This gives
19)
We are now equipped to show that the ancient Sumerians were right in dividing the rotation
period of the Earth (the day) into 24 units (the hour) because
That is
20).
Which is to say the angular momentum of the Earth to its Planck constant gives the base 60
counting in terms of the 24 hour day, the 60 of 60 minutes in an hour, and 60 seconds in a
minute, that determine our base unit of duration we call a second that happens to be, as I have
shown, the base unit of the wave solution to the atom and the Earth/Moon/Sun system.
Our equation in this paper for the Earth energy as a solution of the wave equation (eq. 13)
13).
ϕ
L = 2π M f r
2
L =
4
5
π M f r
2
= 2.8314E 33J s
L
earth
=
7.05E 33
2.8314E 33
= 2.4899 2.5 = 2
1
2
2.5(24) = 60
L
earth
24 = 60
K E
e
= n
R
R
m
G
2
M
2
e
M
3
m
2
2
of 13 53
does not depend on the Moon’s distance from the Earth, only its mass. The Moon slows the
Earth rotation and this in turn expands the Moon’s orbit, so it is getting larger, the Earth loses
energy to the Moon. The Earth day gets longer by 0.0067 hours per million years, and the
Moon’s orbit gets 3.78 cm larger per year. Equation 20
20)
only specifies divide the day into 24 units, and hours into 60 minutes and minutes into 60
seconds, regardless of what the Earth rotational velocity is. But it was more or less the same as it
is now when the Sumerians started civilization. But it may be that it holds for when the Earth
day is such that when the Moon perfectly eclipses the Sun, which we said might be a condition
for the optimization of life preventing extreme hot and cold. That is when 15 holds
15)
Which holds for today and held for the ancient Sumerians and holds for when the Earth rotation
gives the duration of a second we have today.
We want the basis set of equations for the solar system. We have
We have from equation
We have equations
We have equations
L
earth
24 = 60
r
planet
r
moon
R
star
R
moon
λ
moon
=
2
GM
3
m
= 3.0281E8m
λ
moon
c
=
2
GM
3
m
1
c
= 1.0secon d s
= (hC )K E
e
hC = 1secon d
2
3
π r
p
α
4
G m
3
p
1
3
h
c
= 1secon d
(
1
6α
2
4πh
Gc
)
r
p
m
p
= 1secon d
1secon d =
K E
m
K E
e
(Ear th Day)
of 14 53
From these it becomes clear that
21)
22)
23)
Thus combining equation 20 with the following
21).
20).
Which gives
24)
=
This is very accurate to give us a second. But notice 6.262 is approximately . We see that
25)
is the circumference of a unit circle, it could be that the circumference of the Earth orbit, can
be taken as 1 (unity) and essentially we have the mystery of base 60 and the 24 hour day of the
ancient Sumerians is solved, it connects unit circle to 1 (unity).
=
GM
3
m
c
K E
e
K E
e
=
GM
3
m
c
=
(
1
6α
2
r
p
m
p
4πh
Gc
)
K E
e
λ
moon
c
= 1secon d
hC = 1secon d
= (hC )K E
e
K E
e
=
GM
3
m
c
L
earth
24 = 60
1secon d =
(
24
60
)
2
L
2
earth
GM
3
m
c
(
24
60
)
2
(7.05E 33)
2
(6.67408E 11)((7.347673E 22kg)
3
(299792458m /s)
=
(
24
60
)
2
6.262sec = 1.002secon d s = 1.00secon d s
2π
1 =
(
24
60
)
2
2π
2π
of 15 53
26).
27)
I came up with a physical theory for certain aspects of reality that showed that our solar system
had a wave equation solution, where the wave equation was developed for describing atomic
systems, among other things, and is at the heart of quantum physics, the modern physics that
describes things on the micro-scale and our planetary system is on the macro-scale. I found it
was based on the Moon that orbits the Earth, which seems to be some kind of a natural
yardstick, and was further based on the unit of a second for measuring time as the base unit,
which I found turns out to be some kind of a natural constant. This natural constant of a
second also, in the theory, predicts the radius of a proton, the fundamental unit of the atom,
and as such a gap is bridged between the microcosmos and the macrocosmos. Further it
resulted in a description of the hydrocarbons, which are the backbones of the chemistry of life.
Bringing in the biological like this makes it the rudiments of a theory of everything,!
The interesting thing is, though, that the second was not designed, originally, as a natural
constant, but came to us ultimately from the ancient Sumerians of Mesopotamia, who were
some of the rst to settle down from hunting with stone spearpoints to invent agriculture,
mathematics, writing, metallurgy, architecture… to invent civilization.!
The idea was to divide the day into twelve hours, and measure each hour by the position of the
Sun in the sky, as measured by the shadow cast by a rod onto a plate with lines to measure the
time of day. They chose twelve hours in a day because 12 is the smallest abundant number,
which is to say it is evenly divisible by 1,2,3,4,6 whose sum is 16 which is greater than twelve. !
Their counting system was base 60 (sexagesimal) which means they started counting again
after counting to 60 where our system is base 10 (decimal) which means we start counting
again after 10. They chose base 60 because sixty, aside from being an abundant number — its
divisors add up to a value greater than it — it is the rst number divisible by the rst 6 counting
numbers, 1,2,3,4,5,6 aside from being evenly divisible by 10,12,15,30.!
This lead to the ancient Babylonians dividing the hour into 60 minutes, and the minute into 60
seconds. This in the end lead to the ancient Greeks using it for astronomy, and the Italians to in
the Middle Ages build the rst clocks with hours, minutes, and, seconds. In the end they gave
us our clocks and they way we measure time in the West, today. The ancient Egyptians,
separately and independently invented civilization from the Sumerians, but were never invaded
by tribes, like in Sumer, who stayed and adopted their math and writing, but rather were
isolated by desert, ocean, and forests, so as not to spread their invention of Mathematics,
writing, and architecture to the rest of the world, at least until they and the ancient Greeks
interacted and shared knowledges in several exchanges where ancient Greek scholars
wandered into Egypt. But, the ancient Egyptians also divided the day into 12 hours and
invented sun dials to tell time as such. However, we got the 12 hour day, or 24 hour day from
sunrise to sunrise and sunset to sunset, from the ancient Sumerians.!
However, myself, having been born in the West I was more aware of Sumer and Egypt, than
India, and indeed while Sumer and Egypt started civilizations, so did the people who brought it
to India, to the Dravidians, and the idea of transmigration of the soul who were from its
outskirts, who brought it from the Indus Valley where the rivers were, they were the Indus Valley
2
2
=
24
60
π
cos(45
) = cos(π /4) =
24
60
π
of 16 53
Civilization, just as old as Sumer and Egypt. It would seem civilizations always started along
rivers, the Tigris and Euphrates in Mesopotamia, and the Nile in Egypt.!
However, after having looked at my theory for reality in terms of the base 60 of Sumer that gave
us the second, and how its dynamics could have coincided with modern physics, and
suggesting that it was given to them by ancient Aliens, because they said they got their
knowledge from those who came from the heavens, the Anunaki, I began to look at the system
that ancient India used to measure time, and I found that they used base 60 as well, but in a
very reversed way from what we got from Mesopotamia, the Babylonians, and ancient Greeks.
And, I found further, that their system of base 60 describes certain aspects of Nature as well.!
Before the colonization of India by the British, another system of measuring time from ancient
times was used, and it was called the Bharat Clock. There are eorts in India today to return to
this, and they suggest for contemporary physics it serves better, as well as it is more ecient
for productivity in the work force. Further still they explain that it is based on Natural
phenomenon in the Universe, and that in the method of the West, it is not. I would take issue
with that not just because my theory for reality shows the Mesopotamian system gives us the
unit of a second in planetary systems, atomic systems, and biological systems, which they
could not know about, but because it does have an elegance that works well for the the
periods of celestial motions, and has since its inception, though I can see advantages in their
system as well.!
The Hindu system is quite inverted from ours, but my theory of physical reality shows the two
systems are profoundly connected in a way that can be quite revealing. We measure the year
as 365.25 days which is close to the 360 degrees in a circle. As such the Earth moves around
the Sun in its orbit through neatly about 1 degree per day. The Sumerians gave us the 360
degrees circle, because in their base 60 counting 6 times 60 is 360, and 6 is the most dynamic
number, a six-sided regular polygon has its radii equal to its sides, which mathematically has
many advantages for solving problems. It occurs in Nature as the closest packing of equal
radius spheres, the so so called six around one, typical in Nature of petal arrangements around
a center in a ower, and 6-fold symmetry is typical of the physical, like snowakes forming six-
points as ice crystals fall through cold air. 5-fold symmetry is typical of life, like two eyes and
nose and a mouth, or two arms, two legs, and a head.!
The Earth then rotates close to 360 times in a year, close to the time for it to complete an orbit
around the Sun. Each rotation is a day, and from the Sumerians we divided that day into 24
hours. 60 divided by 24 is 2.5 and as it would turn out 2.5 is the factor than connects the Hindu
system to the Western system. As we shall see, my theory nds this factor is a pivotal factor of
Nature. Let’s look at our system and theirs. We have!
(365.25 days)(24 hours)(60 minutes)(60 seconds)=31557600 seconds/year!
And the seconds in a day are!
(24 hours)(60 minutes)(60 seconds)=86400 seconds/day!
Thus we divide the hour into 60 minutes, and the minutes into 60 seconds. The Hindu Vedic
System has a day is 60 ghatika of 24 minutes each, each ghatika is divided into 60 palas of 24
seconds each, and each pala is divided into 60 vispalas, each vispala of 0.4 seconds each. So
where our system has a base unit of 1 second, theirs has a base unit of 0.4 seconds, so that
could be and advantage to their system, a smaller unit of time is more rened. Further their day
is divided into 60 units, ours into only 24, so their hour, the ghatika is only 24 minutes long, and
ours is 60 minutes long, The proponents in India say since when working and doing chores we
of 17 53
do a few chores in an hour, they do about one per ghatika is 24 minutes, which makes the
measure of time more manageable. That could be an advantage I think. They say our hour is so
long because the lines had to be far apart on the Egyptian Sun Dial so the shadow cast by the
Sun didn’t cross-over onto another line. But today, with modern technology we can make clock
lines marking hours closer together and measure them with a pointer hand pointing to them
without any problems, and the result is we would have a smaller more rened hour (60 of them
in a day as opposed to 24). They suggest this method of measuring time would work better in
science and engineering as well, that we have to get away from the way sun dials had to be
made in Egypt in ancient times.!
But they further point out that their system describes Nature. They say theirs are 108,000
vispalas in a day, and 108,000 vispalas in a night giving 216,000 vispalas in a 24 hour day. The
diameter of the Sun is 108 that of the Earth, and the average distance from the Sun to the
Earth is 108 solar diameters, and the average distance from the Moon to Earth is 108 lunar
diameters. 108(10)(10)(10)=108,000. Ourselves and them use base 10 counting, and that is
probably because we have ten ngers to count on.!
Conclusion
In conclusion we have that the unit of a second as derived from the Mesopotamians is a Natural
constant. As well we see the Hindu Vedic system describes Nature as well. It says
( )
( )
( )
( )
To see how this is connected to nature we would have to go into the significance of base 10
counting as we did with base 60 counting. But interesting here is that
2R
= 2(6.957E8m) = 1.3914E 9m eters
R
= SolarRadiu s
r
e
= 1.496E11m
r
e
= Ear thOrbit
r
e
2R
= 107.52 108
2R
moon
= 2(1.74E6m) = 3.48E6m
R
moon
= Lun arRa dius
r
moon
= 3.845E8m
r
moon
= Lun arOrbit
r
m
2R
moon
= 110 108
(86,000sec /d a y)
2
=
43200
0.4sec /vispa l a
= 108,000vispalas /d a y
108(10)(10)(10) = 108,000
r
e
2R
= 107.52 108
r
m
2R
moon
= 110 108
of 18 53
because
13.
where
15.
16.
Which is to say radius of the Sun to the radius of the Moon in the planetary equation plays the
role of the number of protons (Z) in an element for the energy of the electron orbits in the
atomic equation and that the Moon perfectly eclipses the Sun as see from the earth (eq 15) might
be a condition for sophisticated habitable star systems, we know the Moon allows for the
seasons making life very successful on Earth by preventing temperature extremes. So in the
Hindu Vedic system the diameters of the Sun and Moon fitting into Earth orbit and Lunar orbit
the same amount might play a similar role in a theory. You would have to go into base 10 like we
did with base 60. We have
West
24 hours
60 minutes
60 seconds
India
60 hours
24 minutes
24 seconds
A sort of inversion of things. The incredible thing here is that the wonderful equations,
equations 19 and 20:
19.
20.
unify the Mesopotamian system with the Hindu system because
R
R
e
=
6.957E8m
6,371,000m
= 109 108
K E
e
= n
R
R
m
G
2
M
2
e
M
3
m
2
2
r
planet
r
moon
R
star
R
moon
R
R
m
=
6.96E8m
1737400m
= 400.5986
L
earth
=
7.05E 33
2.8314E 33
= 2.4899 2.5 = 2
1
2
L
earth
24 = 60
of 19 53
This suggests that the Hindu system is just as Natural as the Western system.
Equation 25 that says
25)
1 hour=(60min/hour)(60sec/min)=3600sec=1 Western Hour
1 vispala=0.4sec
60 vispala=(60)(0.4sec)=24 sec=1pala
60 pala = (60)(24 sec)=1440 sec=24 min = 1 ghaticka = 1 HinduHour
28).
where is the circumference of a unit circle, and we have unity on the left, so it could be if you
combine the Indian system of measuring time with the Western system you would have a system
of units for the solar system where the equations work out nicely.
1vi spala = 0.4secon d s
(0.4s)(2.5) = 1secon d
1 =
(
24
60
)
2
2π
1 =
(1440sec)
2
(3600sec)
2
2π
1 =
(1gh at ick a)
2
(1h our)
2
2π
1 =
Hin du Hour
2
Wester n Hour
2
2π
1 =
(60pala)
2
(60min)
2
2π =
(24min)
2
(60min)
2
2π
1 =
((60)(24sec))
2
((60)(60sec))
2
2π =
(60pala)
2
(60sec)
2
2π =
(24sec)
2
(60sec)
2
2π
2π
of 20 53
Part 2: Biological Life Part of a Universal Natural Process In The Universe
of 21 53
Introduction I have a theory for the star systems, the planets and their suns, wherein such
systems are solved with the Schrödinger wave equation that is used to solve atomic systems. The
result is that star systems and atomic systems, systems on the macro scales and micro scales, are
governed by the same underlying principles. It came to pass that this theory indicated an
overlap with biological systems indicating that biological life could be part of the same idea, and
that is what I hope to pull out of that paper and develop here.
The theory for the planets, which also predicted the radius of a proton, which is one of the
fundamental particles out of which matter is made, suggested that star systems which support
life are a part of a Universal natural process. One of the conditions for star systems that support
life made use of the interesting fact that has mystified science for some time now, that the Moon
is the right size and distance from the Earth that it nearly perfectly eclipses the Sun. The reason
a moon would be necessary for life to be optimally successful on a habitable planet is that we
know our moon that orbits our Earth, makes life here successful because it holds the Earth at its
inclination to its orbit around the Sun allowing for the seasons, thus preventing extreme hot and
extreme cold.
Interestingly, the solution for the planets of the Schrödinger wave equation, was quantized in
terms of the Earth’s moon, and a base unit of one second. The Moon seems to be some sort of a
natural yardstick. The strange thing is that the basis unit of time is one second and the second
was not developed with the planets and atoms in mind, so it is strange that it lands upon the
function of a natural constant in our theory. In the theory (An Abstract Theory of Reality,
Beardsley 2024) I talk a little about why that might be. The second came to us from the Ancient
Sumerians, who invented civilization with settling down from following the herds and hunting
with stone spearpoint to invent agriculture, metallurgy, writing, and mathematics. They had a
base 60 counting system, and found it convenient then to divide the Earth day (rotation period)
into 24 hours, which in the end became adopted by the world. Their base 60 counting resulted in
the Babylonians dividing the hour into 60 minutes, and that in turn into 60 seconds, and that is
how we have the duration of a second we have today. They chose base 60 (sexagesimal) because
60 is evenly divisible by so much from which they gave us the 12 hour day and 12 hour night or
24 hour day from sunrise to sunrise, sunset to sunset. 12 times 5 is 60, also 60 times 6 is 360,
they gave us the 360 degree circle as well.
In this paper I show that the same unit of a second that describes planetary systems and atomic
system in common describes hydrocarbons, the skeletons of biological life chemistry. I further
show that it predicts the atomic radii of the hydrogen and carbon atoms from which such
skeletons are made. Hydrogen and carbon are the most abundant elements in life chemistry,
carbon is the core element upon which life is made, and hydrogen is the simplest element,
element one in the periodic table of the elements, consisting of 1 proton, and is the most
abundant element in the universe by far, and plays the dominant role in all of chemistry.
of 22 53
A Theory For Biological Hydrocarbons I have found that the basis unit of one second is
not just a Natural constant for physical systems like the atom and the planets around the Sun
(See my paper An Abstract Theory of Reality, Beardsley 2024) but for the basis of biological
life, that it is in the sixfold Nature of the chemical skeletons from which life is built, the
hydrocarbons. I found
1).
2).
Where is the fine structure constant, is the radius of a proton, is the mass of a
proton, is Planck’s constant, is the constant of gravitation, is the speed of light, is the
kinetic energy of the Moon, is the kinetic energy of the Earth, and is the
rotation period of the Earth is one day.
The first can be written:
3).
4).
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting the units of cancel with the bodies of these equations on the
left, which are in units of mass, but rather divide into them, giving us a number of protons. I say
this is the biological because these are the hydrocarbons the backbones of biological chemistry.
We see they display sixfold symmetry. I can generate integer numbers of protons from the time
values from these equations for all of the elements with a computer program. Some results are:
1secon d =
1
6α
2
r
p
m
p
4πh
Gc
1secon d =
K E
m
K E
e
(Ear th Day)
α = 1/137
r
p
m
p
h
G
c
K E
m
K E
e
Ear th Day
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton 6secon d s = hydrogen(H )
m
p
of 23 53
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. And carbon is the core element of life. We see the
duration of the base unit of measuring time, 1 second, given to us by the ancients (the base 60,
sexagesimal, system of counting of the Sumerians who invented math and writing and started
civilization), is perfect for the mathematical formulation of life chemistry. Here is the code for
the program, it finds integer solutions for time values, incremented by the program at the
discretion of the user:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792458,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
We have that
Since this 6 seconds is also proton-seconds we have
5). is carbon (C)
6). is hydrogen (H)
1
α
2
r
p
m
p
4πh
G c
= 18769
0.833E 15
1.67262E 27
4π (6.62607E 34)
(6.67408E 11)(299,792458)
= 6.029978s 6s
1
6pr ot on s
1
α
2
r
p
m
p
4πh
G c
= 1secon d
1
1pr ot on
1
α
2
r
p
m
p
4πh
G c
= 6secon d s
of 24 53
Our Theory For Hydrocarbons With all that has been said we are equipped to proceed. We
want to consider the radius of a hydrogen atom and the radius of a carbon atom. The radius of
a carbon atom given in your periodic table of the elements is often 70 to 76 picometers. The
covalent radius of hydrogen is given as 31 picometers. The atomic radius of hydrogen is 53
picometers and the atomic radius of carbon is 67 picometers. We want to consider the atomic
radii of both, because the covalent radius, determined by x-ray diraction for diatomic
hydrogen, is the size of two hydrogen atoms joined H2 divided by two, where it is measured
that way, joined, in the laboratory. Carbon is C2 divided by 2. We are interested in the single
carbon and hydrogen atoms, because we want to know what our theory for their six-fold
symmetry with one another in their representations in proton-seconds says about the way they
combine as the skeletons of life chemistry. We start with the Planck constant, which is like
ux, a mass (perhaps a number of particles) per second over an area. That is it is kilograms per
second over square area:!
!
We have equations 1 and 2:!
1). is carbon (C)
2). is hydrogen (H)
We can write these
=
3).
=
4). !
We have from 3 and 4…!
5). !
We nd the ratio between the surface areas of the hydrogen and carbon atoms:!
h,
h = 6.62607E 34J s = 6.62607E 37
kg
s
m
2
1
6pr ot on s
1
α
2
r
p
m
p
4πh
G c
= 1secon d
1
1pr ot on
1
α
2
r
p
m
p
4πh
G c
= 6secon d s
(
6.62607E 37
kg
s
m
2
)
6secon d s
m
p
(
6.62607E 37
kg
s
m
2
)
6seconds
1.67262E 27kg
= 2.37689E 6m
2
(
6.62607E 37
kg
s
m
2
)
1second
6m
p
(
6.62607E 37
kg
s
m
2
)
1second
6(1.67262E 27kg)
= 6.602486E 8m
2
h
m
p
(6seconds)
1
h
m
p
(1second)
6
=
2.37689E 6m
2
6.602486E 8m
2
= 35.9999 40
of 25 53
6). !
7). !
!
It is the golden mean. These atomic radii are the radii between the nucleus of the atoms and
their valence shell, which is what we want because the valence shell is the outermost electrons
responsible for the way the hydrogen and the carbon combine to make hydrocarbons. We will
write this!
8). !
I am guessing the reason we have the golden mean here is that it is the number used for
closest packing. But what we really want to do is look at the concept of action, for hydrogen
given by six seconds and carbon given by 1 second. We take equations 1 and 2:!
1). is carbon (C)
2). is hydrogen (H)
And, we write
9).
Where is the radius of the atom, and t its time values given here by equations 1 and 2. We have for
hydrogen
10).
=
This is actually very close to the radius of a hydrogen which can vary around this depending on how you
are looking at it, which we said is given by 5.3E-11m. For carbon we have:
11).
=
And this is actually very close to the radius of a carbon atom which is 6.7E-11m. The thing is if we consider
the bond length of the simplest hydrocarbon CH4, methane, which can be thought of as
H
surface
= 4π (r
2
H
) = 3.52989E 20m
2
C
surface
= 4π (r
2
C
) = 5.64104E 20m
2
4π(53pm)
2
4π(67pm)
2
= 0.62575 0.618... = ϕ
r
2
H
r
2
C
= ϕ
HC
1
6pr ot on s
1
α
2
r
p
m
p
4πh
G c
= 1secon d
1
1pr ot on
1
α
2
r
p
m
p
4πh
G c
= 6secon d s
m
p
G c
h
t
0
dt = r
A
r
A
m
p
G c
h
t
0
dt = (1.67262E 27k g)
(6.67408E 11)(299,792,458)
6.62607E 34
6sec
0
dt
(9.2E 12m /s)(6secon d s) = 5.5E 11m
m
p
G c
h
t
0
dt = (1.67262E 27k g)
(6.67408E 11)(299,792,458)
6.62607E 34
6sec+1sec
0
dt
(9.2E 12m /s)(7secon d s) = 6.44E 11m
of 26 53
16).
our, equations give
17).
which are the same thing. Thus we have the basis for a theory of everything in that it includes the macro
scale, the Earth/Moon/Sun System, because we had:
and it includes the radius of the proton which is the radius and mass of a proton and gives 1
second
and we have the hydrocarbons the skeleton of the chemistry of life give one second
1). is carbon (C)
2). is hydrogen (H)
and because they predict the radii of the carbon and hydrogen atoms at the core of life
10).
=
11).
=
We want to proceed with the elements nitrogen (N) and oxygen (O). We just did carbon and hydrogen, but
hydrocarbons are then combined with primarily nitrogen and oxygen to make amino acids, the building
blocks of life. DNA synthesizes them into proteins, for hair, muscles tissue, skin, nails, etc The most
basic organic compound, HNCO (isocyanic acid) is just one each of these elements. It bonds
H-N=C=O
It was shown by the Miller-Urey experiment that if a primordial Earth was mainly ammonia (NH3),
Methane (CH4), water (H2O) these combined under energy form 11 of the 20 biological amino acids life
needs for DNA to synthesize proteins. Thus we start with nitrogen and notice the first eight element are
atomic number 1 is hydrogen, atomic number 2 is helium, atomic number 3 is lithium, atomic number 4 is
beryllium, atomic number 5 is boron, atomic number six is carbon, atomic number 7 is nitrogen, atomic
number 8 is oxygen. The atomic numbers correspond to the the number of protons in each element.
r
H
+ r
C
= 5.3E 11m + 6.7E 11m = 1.2E 10m
r
H
+ r
C
= 5.5E 11m + 6.44E 11m = 1.2E 10m
1secon d =
K E
m
K E
e
(Ear th Day)
1secon d =
1
6α
2
r
p
m
p
4πh
Gc
1
6pr ot on s
1
α
2
r
p
m
p
4πh
G c
= 1secon d
1
1pr ot on
1
α
2
r
p
m
p
4πh
G c
= 6secon d s
m
p
G c
h
t
0
dt = (1.67262E 27k g)
(6.67408E 11)(299,792,458)
6.62607E 34
6sec
0
dt
(9.2E 12m /s)(6secon d s) = 5.5E 11m
m
p
G c
h
t
0
dt = (1.67262E 27k g)
(6.67408E 11)(299,792,458)
6.62607E 34
6sec+1sec
0
dt
(9.2E 12m /s)(7secon d s) = 6.44E 11m
of 27 53
We suggest the periodic table of the elements has its function in sixfold symmetry because it has
18 groups in that 6 times 3 is 18 and 3 this the first odd prime number other than one, where all
prime numbers are odd except 2, and 1 defines a prime number in that a prime number is a
number that is evenly divisible by itself and 1. The periodic table is then centered around
hydrogen and carbon, hydrogen with one proton, carbon with six where hydrogen is given by 6
seconds and carbon is given by 1 second, all values beyond 6 seconds are fractional protons and
1 second is the smallest integer number of seconds that gives whole protons. That is, equations 1
and 2 are
1). is carbon (C)
2). is hydrogen (H)
We already had that hydrogen, the base element with 1 proton gives itself, hydrogen, and carbon with 6
protons is given by itself and hydrogen in equations 10 and 11:
10).
=
11).
=
We had that experimentally for the hydrocarbon backbones of life
16).
And, according to our theory that
17).
We want to consider that elements 6, 7, 8 which are carbon, nitrogen, and oxygen, respectively then
correspond to H, He, Li, and Be element 1, 2, 3, and 4. Carbon (6) already corresponds to hydrogen (H),
so nitrogen (7) corresponds to H, He, Li, and oxygen (8) corresponds to H, He, Be. This also works
because nitrogen (group 15) wants to have stable noble gas electron configuration (group 18) and 18-15=3
meaning it ionizes and oxygen (group 16) wants the same so 18-16=4 meaning it oxidizes . Thus
1
6pr ot on s
1
α
2
r
p
m
p
4πh
G c
= 1secon d
1
1pr ot on
1
α
2
r
p
m
p
4πh
G c
= 6secon d s
m
p
G c
h
t
0
dt = (1.67262E 27k g)
(6.67408E 11)(299,792,458)
6.62607E 34
6sec
0
dt
(9.2E 12m /s)(6secon d s) = 5.5E 11m
m
p
G c
h
t
0
dt = (1.67262E 27k g)
(6.67408E 11)(299,792,458)
6.62607E 34
6sec+1sec
0
dt
(9.2E 12m /s)(7secon d s) = 6.44E 11m
r
H
+ r
C
= 5.3E 11m + 6.7E 11m = 1.2E 10m
r
H
+ r
C
= 5.5E 11m + 6.44E 11m = 1.2E 10m
O
2
of 28 53
lithium is 3 protons, and He+H=3 protons for nitrogen ( ) and oxygen ( ) is Be-He=2. We have
that
We have
Nitrogen
Oxygen
Thus from equation 9
Nitrogen
9)
=
Where experimentally
And, theoretically
So we are in agreement.
N
3
O
2
1
6pr ot on s
1
α
2
r
p
m
p
4πh
G c
=
1
6pr ot on s
(6pr ot on s secon d s) = 1secon d = C
1
1pr ot on s
1
α
2
r
p
m
p
4πh
G c
=
1
1pr ot on s
(6pr ot on s secon d s) = 6secon d s = H
1
2pr oton s
1
α
2
r
p
m
p
4πh
G c
=
1
2pr oton s
(6pr ot on s secon d s) = 3secon d s = He
1
3pr ot on s
1
α
2
r
p
m
p
4πh
G c
=
1
3pr ot on s
(6pr ot on s secon d s) = 2 seco n d s = Li
1
2pr oton s
1
α
2
r
p
m
p
4πh
G c
=
1
2pr oton s
(6pr ot on s secon d s) = 3secon d s = He
1
4pr ot on s
1
α
2
r
p
m
p
4πh
G c
=
1
4pr ot on s
(6pr ot on s secon d s) = 1.5secon d = Be
m
p
G c
h
t
0
dt = r
A
N
3
= m
p
G c
h
t
0
dt = (9.2E 12m /s)
1sec+2sec+3sec
0
dt
5.52E 11m
NH
3
= r
N
+ r
H
= 5.6E 11m + 5.3E 11m = 1.1E 10m
NH
3
= r
H
+ r
H
= 5.52E 11m + 5.5E 11m = 1.1E 10m
of 29 53
Oxygen
=
Where experimentally
And, theoretically
So we are in good agreement.
O
2
= m
p
G c
h
t
0
dt = (9.2E 12m /s)
1sec+1.5sec+3sec
0
dt
5.06E 11m
H
2
O = r
O
+ r
H
= 4.8E 11m + 5.3E 11m = 1.01E 10m
H
2
O = r
O
+ r
H
= 5.06E 11m + 5.5E 11m = 1.056E 10m
of 30 53
Part 3: The Solar Formulation For The Solution Of The Wave Equation
of 31 53
The Solar Formulation
Our solution of the wave equation for the planets gives the kinetic energy of the Earth from the
mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting the
Sun. In our lunar formulation we had:
1.
We remember the Moon perfectly eclipses the Sun which is to say
2.
Thus equation 1 becomes
3.
The kinetic energy of the Earth is
4.
Putting this in equation 3 gives the mass of the Sun:
5.
We recognize that the orbital velocity of the Moon is
6.
So equation 5 becomes
7.
This gives the mass of the Moon is
8.
Putting this in equation 1 yields
9.
K E
e
= 3
R
R
m
G
2
M
2
e
M
3
m
2
2
R
R
m
=
r
e
r
m
K E
e
= 3
r
e
r
m
G
2
M
2
e
M
3
m
2
2
K E
e
=
1
2
GM
M
e
r
e
M
= 3r
2
e
GM
e
r
m
M
3
m
2
v
2
m
=
GM
e
r
m
M
= 3r
2
e
v
2
m
M
3
m
2
M
3
m
=
M
2
3r
2
e
v
2
m
K E
e
=
R
R
m
G
2
M
2
e
M
2r
2
e
v
2
m
of 32 53
We now multiply through by and we have
10.
Thus the Planck constant for the Sun, , in this the case the star is the Sun, is angular
momentum quantized, the angular momentum we will call , the subscript for Planck. We
have
We write for the solution of the Earth/Sun system:
11.
Let us compare this to that of an atom:
12.
We notice that in equation 11
; ; ; ;
is really . We can write 11 as
13.
We say. . That is
Let us see how accurate our equation is:
M
2
e
/M
2
e
K E
e
=
R
R
m
G
2
M
4
e
M
2r
2
e
v
2
m
M
2
e
L
p
p
L
p
= r
e
v
m
M
e
= r
e
v
m
M
e
= (1.496E11m)(1022m /s)(5.972E 24kg) = 9.13E 38kg
m
2
s
L
2
p
= r
2
e
v
2
m
M
2
e
= 7.4483E 77J m
2
kg = 8.3367E 77kg
2
m
4
s
2
K E
e
=
R
R
m
G
2
M
4
e
M
2L
2
p
E =
Z
2
n
2
k
2
e
e
4
m
e
2
2
Z
2
n
2
R
R
m
k
2
e
G
2
e
4
M
4
e
m
e
M
2
L
2
p
L
p
K E
e
=
R
R
m
G
2
M
4
e
M
2
2
=
/2π
= 9.13E 38J s
h
= 2π
= 5.7365E 39J s
K E
e
=
R
R
m
G
2
M
4
e
M
2
2
of 33 53
=
=
We have that the kinetic energy of the Earth is
Our equation has an accuracy of
Which is very good.
We call the same in our solar solution. But we now want our solution for the solar
formulation. is the mass of the Earth and is the mass of the Sun. We have our solution
might be!
14. !
Where is !
We have
15.
This has an accuracy of
!
Thus the solutions to the wave equation!
R
R
m
(6.67408E 11)
2
(5.972E 24kg)
4
(1.9891E 30kg)
2(8.3367E 77kg
2
m
4
s
2
)
R
R
m
(6.759E 30J )
R
R
m
=
6.957E8m
1737400m
= 400.426
K E
e
= 2.70655E 33J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.70655E 33J
2.7396E 33J
= 98.79 %
L
p
r
n
M
e
M
r
n
=
2
GM
3
e
R
m
R
h
= 9.13E 38J s
r
3
=
(9.13E 38)
2
(6.67408E 11)(5.972E 24kg)
3
1
400.5986
= 1.4638E11m
1.4638E11m
1.496E11m
100 = 97.85 %
of 34 53
!
for the solar formulations are!
13)
14)
Equating The Lunar And Solar Formulations Yield Our 1 Second Base Unit
Let us equate equation 1 with equation 13:
1.
13.
This gives:
16.
We remember that
2
2m
[
1
r
2
r
(
r
2
r
)
+
1
r
2
sinθ
θ
(
sinθ
θ
)
+
1
r
2
sin
2
θ
2
ϕ
2
]
ψ + V(r)ψ = E ψ
K E
e
=
R
R
m
G
2
M
4
e
M
2
2
r
n
=
2
GM
3
e
R
m
R
= 9.13E 38J s
h
= 2π
= 5.7365E 39J s
K E
e
= n
R
R
m
G
2
M
2
e
M
3
m
2
2
K E
e
=
R
R
m
G
2
M
4
e
M
2
2
3
R
R
m
G
2
M
2
e
M
3
m
2
2
=
R
R
m
G
2
M
4
e
M
2L
2
p
L
p
=
M
2
e
M
M
3
m
3
= (hC )K E
p
hC = 1secon d
C =
1
3
1
α
2
c
1
3
2π r
p
G m
3
p
of 35 53
This gives
17.
We have
18.
This equates the orbital velocity of the Moon with the centripetal acceleration of the Earth in
terms of one second by way of the mass of the Earth, the mass of the Sun, the mass of the Moon,
and the orbital number of the Earth. Let us compute
19.
Let us see how well equation 18 works. at aphelion is 966 m/s and .
. We have
That is an accuracy of
Equation 18 can be written:
20.
From our equation:
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
r
e
v
m
M
e
=
1
6α
2
r
p
m
p
h 4π
Gc
1
2
M
e
v
2
e
M
2
e
M
M
3
m
3
2v
m
=
v
2
e
r
e
(1secon d )
M
2
e
M
M
3
m
3
M
2
e
M
M
3
m
3
=
(5.972E 24kg)
2
(1.9891E 30kg)
(7.34763E 22kg)
3
(1.732)
= 321,331.459 321,331
v
m
v
e
= 29,800m /s
r
e
= 1AU = 1.496E11m
2(966m /s) =
(29,800m /s)
2
1.496E11m
(1sec)(321,331.459)
1,907m /s = 1,932m /s
1907
1932
= 98.7 %
1secon d = 2r
e
v
m
v
2
e
M
3
m
3
M
2
e
M
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
of 36 53
We have
21.
Since , the diameter of the Earth orbit, we have
22.
And we see the Earth/Moon/Sun system determines the radius and mass of the proton and vice
versa. We basically have
23. 62)
Where is the Earth orbital number. We have
= 7.83436E4seconds
EarthDay=(24)(60)(60)=86400 seconds,
accuracy
This last equation, equation 62, we can use to find the rotation period, or the length of the day,
of an earth-like planet in the habitable zone of any star system, so it would be very useful.
We want to turn our attention to equation 20 and write it
1
6α
2
r
p
m
p
h 4π
Gc
= 2r
e
v
m
v
2
e
M
3
m
3
M
2
e
M
2r
e
= d
e
1
6α
2
r
p
m
p
h 4π
Gc
= d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
1secon d = 2v
m
r
e
v
2
e
M
3
m
3
M
2
e
M
1secon d =
K E
moon
K E
earth
Ear th Day
1secon d =
M
m
v
2
m
M
e
v
2
m
Ear th Day
Ear th Day =
2r
e
v
m
M
m
M
n
n = 3
Ear th Day =
2(1.496E11m)
966m /s
7.34763E 22kg
1.9891E 30kg
1.732
7.834E4s
86,400s
100 = 90.675 %
of 37 53
24.
We see equating solar and lunar formulations for energy yield the base unit of one second.
Equating the lunar and solar solutions for orbitals instead of the for energies, which we just did,
we have
Yields
25.
Where . The accuracy of this is
Thus the energy equations gave the equation 55:
16. !
And equating the orbital equations gives
26.
These last two yield
27.
The accuracy is
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
2
2
GM
3
m
R
R
m
1
n
=
L
2
p
GM
3
e
R
m
R
2
M
3
e
M
3
m
=
3
2
L
2
p
R
2
m
R
2
s
3/2 = cos(π /6)
(2.8314E 33)
2
(5.972E 24)
3
(7.34763E 22)
3
= (0.866)(9.13E 38)
2
(1737400)
2
(6.96E8)
2
4.29059E 72 = 4.5005E 72
4.29059E 72
4.5005E 72
100 = 95.3 %
L
p
=
M
2
e
M
M
3
m
3
L
2
p
=
2 3
3
M
3
e
M
3
m
R
2
R
2
m
2
2
R
R
m
M
e
M
= 1
of 38 53
Is 98% accuracy. The important thing that comes out of this is our base unit of a second because
we see it is also a function lunar, solar, and earth masses.
24.
(400.5986)
5.972E 24kg
1.9891E 30kg
= 1
0.98 = 1
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
of 39 53
Part 4: Wave Solutions For Jupiter And Saturn
of 40 53
Solutions For Jupiter And Saturn
We have the solution for the Bohr hydrogen atom is
1
is the atomic number, the number of protons orbited, but for our system it is 1, because one
body is orbited. is the orbit number, which we will deal with later. We have our solution for the
Earth/Moon/Sun system is
2
Let is the earth mass, is the lunar mass, and is our
Planck constant for the Earth/Moon/Sun system. We compute equation 2:
=
The kinetic energy of the Earth is
This is
The Moon perfectly eclipses the Sun which means as seen from the Earth the Moon appears to
be the same size as the Sun. This is because
3
Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,
and is the Moon’s radius. This is because though the Moon is 400 times further from the Sun
than it is from the Earth, the Sun is 400 times larger than the Moon. That is
Thus
4
We find the quantum mechanical solution to the Earth/Moon/System is
E =
Z
2
(k
e
e
2
)
2
m
e
2
2
n
2
Z
n
E =
G
2
M
2
m
3
2h
2
n
2
M = M
e
m = M
m
h
= 2.8314E 33J s
E =
(6.67408E 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
n
2
3.93E 30Joules
n
2
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
1/n
2
= 697
r
e
r
m
R
R
m
r
e
r
m
R
R
m
6.96E8m
1737400m
= 400.5986
R
R
m
= 400
of 41 53
5 !
for Earth orbit (Third Planet).!
!
!
We see the solar Planck constant we developed works in solving the Schrödinger wave equation
for the Earth/Moon/Sun system. My belief as to why this was never done is that it had to be
realized that the solution of the Earth kinetic energy around the Sun as a quantum mechanical
state is based on the Moon around the Earth.
Now we want to find what the wave equation solutions are for Jupiter and Saturn because they
significantly carry the majority of the mass of the solar system and thus should embody most
clearly the dynamics of the wave solution to the Solar System.
I find that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid,
and go to the gas giants and ice giants, the atomic number is no longer squared and the square
root of the the orbital number moves from the numerator to the denominator. I believe this is
because the solar system here should be modeled in two parts, just as it is in theories of solar
system formation because there is a force other than just gravity of the Sun at work, which is the
radiation pressure of the Sun, which is what separates it into two parts, the terrestrial planets on
this side of the asteroid belt and the giants on the other side of the asteroid belt. The effect the
radiation pressure has is to blow the lighter elements out beyond the asteroid belt when the
solar system forms, which are gases such as hydrogen and helium, while the heavier elements
are too heavy to be blown out from the inside of the asteroid belt, allowing for the formation of
the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the atomic
number of the heavier metals such as calcium for the Earth, while the equation for the giants has
the atomic numbers of the gasses. We write for these planets
6
So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):
E = n
R
R
m
G
2
M
2
m
3
2h
2
n = 3
3(400.5986)(3.93E30J ) = 2.7268585E 33J
2.7268585E33
2.7396E33
= 99.5 %
E =
Z
n
G
2
M
2
m
3
2
2
K E
j
=
1
2
(1.89813E 27kg)(13720m /s)
2
= 1.7865E 35J
E =
Z
H
5
(6.67408E 11)
2
(1.89813E 27kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
E =
Z
H
5
(3.971E 35J ) = Z
H
(1.776E 35J )
of 42 53
Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that
.
Our equation for Jupiter is
7
Where is the atomic number of hydrogen which is 1 proton, and for the orbital
number of Jupiter, . Now we move on to Saturn…
=
The equation for Saturn is then
8
It makes sense that Saturn would use Helium in the equation because Saturn is the next planet
after Jupiter and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well,
just like Jupiter, Saturn is primarily composed of hydrogen and helium gas.
Our equation in this paper for the Earth orbit does not depend on the Moon’s distance from the
Earth, only its mass. The Moon slows the Earth rotation and this in turn expands the Moon’s
orbit, so it is getting larger, the Earth loses energy to the Moon. The Earth day gets longer by
0.0067 hours per million years, and the Moon’s orbit gets 3.78 cm larger per year. It is believed
the Moon came from a chunk knocked off the Earth from a collision with a Mars sized
protoplanet. At the time that the Moon formed it was about 4 Earth radii distant 4.5 billion
years ago. After 100 million years the Moon became tidally locked making its rotation period
equal to its orbital period keeping one face always towards the Earth. After 500 million years the
Moon was orbiting at about 20 Earth radii. It is believed that during the period of heavy
bombardment in a chaotic early solar system about 4.1 to 3.8 billion years ago a large object did
a close pass pulling the Moon further changing its orbit giving it its 5 degree offset from the
Earth’s equator. Our wave equation solution may only use the Moon’s mass but the equation for
Z
H
=
1.7865E 35J
1.776E 35J
= 1.006proton s 1.0proton s = hydrogen(H )
Z = Z
H
E =
Z
H
5
G
2
M
2
j
M
3
m
2
2
Z
H
n = 5
n = 5
K E
S
=
1
2
(5.683E 26kg)(10140m /s)
2
= 2.92E 34J
E =
Z
6
(6.67408E 11)
2
(5.683E 26kg)
2
(7.347673E 22)
3
2(2.8314E 33)
2
Z
2.45
(3.5588E 34J ) = Z(1.45259E 34J )
Z(1.45259E 34J ) = (2.92E 34J )
Z = 2protons = Helium(He)
E =
Z
He
6
G
2
M
2
s
M
3
m
2
2
of 43 53
kinetic energy of the Moon to kinetic energy of the Earth times the Earth Day equal to about one
second:
Which we connect with the equation where the proton gives one second
This holds for when the Moon was at a distance from the Earth such that it appears the same
size as the Sun, which means:
Which is when the two equations above for one second connect to our wave equation solution to
the Earth
Because .
The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the
Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get too
extremely hot or too extremely cold. We see the Moon may be there for a reason.
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996354secon d s
r
planet
r
moon
R
star
R
moon
K E
e
= n
R
R
m
G
2
M
2
e
M
3
m
2
2
R
star
R
moon
=
R
R
m
of 44 53
Part 5: Modeling Habitable Star Systems
of 45 53
Modeling Habitable Star Systems
We want to solve a habitable planetary system in general and apply it to another star system. We
must include the solution of its moon as well, because the Moon of the Earth makes life
optimally possible. However, our Moon is counter-intuitive, it breaks all of the rules of the other
moons in our solar system. One is it has a low mass for its size. It has been said that it could be
hollow. Even suggested that it is a hollow-spacecraft put there to make life as we know it
possible. In our quantum theory for the solar system, we may have discovered the science of the
engineers who may have made the Moon, and we are possibly applying their science here. Our
quantum theory for the solar system also has a solar solution as compared to a lunar solution, so
it works for star systems in general.
The perfect star system to use for our purposes here would be Tau Ceti. Tau Ceti is a star
spectrally similar to the Sun and the closest of such a G-class star, at about 12 light years distant.
Its mass is 78% of the Sun’s. Its mass is
Its radius is:
And its luminosity is
It has several unconfirmed planets with one potentially in the habitable zone. It has been the
subject of science fiction literature. It is in the constellation Cetus. From Tau Ceti the Earth
would be seen to be in the northern hemisphere constellation Bootes with an apparent
magnitude of 2.6. It is spectral class G8 V where our Sun is G2 V.
For the luminosity of a main sequence star we only need to know the mass of the star to get its
luminosity:
1
This holds in solar masses and solar luminosities. The habitable zone of a star is given by
luminosity. If the star is 100 times brighter than our Sun, then the habitable zone is 10 times
further than the Earth is from the Sun by the inverse square law.
2
In astronomical units and solar luminosities. From we can get the orbital period in Earth
years:
3
We can get the radius of the Moon of the planet if we know the mass of the planet, :
M
s
= 0.783 + / 0.012M
R
s
= 0.793 + / 0.004R
L
s
= 0.488 + / 0.0010L
L = M
3.5
r
p
= L
s
r
p
T
2
p
= r
3
p
M
e
of 46 53
4
We want the mass of the Earth from the mass of the Sun. We would guess we can get it from the
mass of the Sun, the size of the Sun, and the base 60 sexagesimal system upon which we have
found everything is based. I find the relationship exists for the Earth:
5
(36)(5)=180=360/2 and 360/6=60 is base 60 sexagesimal. This gives
This is an accuracy of
We can get the orbital distance of the planet’s Moon from the planet:
6
Once we have that we can get the radius of the Moon of the planet
7
Now we can get the rotation period of the planet, its length of day:
8
Because the orbital velocity of the planets Moon is
9
We can use the the delocalization time get the moons mass, which is 1/2 the planets year, with
10 , See Appendix 1
2
R
R
m
M
e
M
= 1
M
e
=
5
6
2
M
R
2
r
2
e
M
e
= 0.13889(1.9891E 30kg)
6.96E8m
2
1.496E11m
2
= 5.979748E 24kg
5.972E 24
5.979748E 24
100 = 99.87 %
r
planet
r
moon
R
star
R
moon
2
R
R
m
M
e
M
= 1
Ear th Day =
2r
e
v
m
M
m
M
3
v
m
=
GM
e
r
m
τ =
m
moon
(2r
moon
)
2
s
of 47 53
And the whole system is solved. Let us see how the theoretical luminosity appears with that
measured with equation 8.1:
It is close to the measured. The equation for predicting is approximate as the actual
values can vary a little with things like metallicity of the stars. Tau Ceti has low metallicity
compared to our Sun.
Lets find the orbital distance of the habitable planet if it exists with equation 2:
Lets find the orbital period of the habitable planet (Its year) with equation 3:
,
=
Lets get the mass of the habitable planet from equation 5:
=
Which is good because the planet, if it exists, is thought to be larger than the Earth. Lets get the
radius of its moon from equation 4:
=
Lets see at what distance it orbits the planet with equation 6:
L
s
= M
3.5
s
= 0.783
3.5
= 0.425L
0.448L
r
p
= 0.488 = 0.69857AU = (0.6986AU )(1.496E11m /AU ) = 1.045E11meters
T
2
p
= r
3
p
T
2
p
= (0.69867AU )
3
T
p
= 0.584years(365.25d a ys)((24hrs)(60min)(60sec) = 1849638secon d s
(0.584)(365.25d ays) = 213.306d a ys
M
p
=
5
36
M
s
R
2
s
r
2
p
M
s
= (0.783)(1.9891E 30kg) = 1.5575E30kg
R
s
= (0.793)(6.96E8m) = 5.52E 8m
M
p
=
5
36
(1.5575E 30kg)
(5.52E8m)
2
(1.045E11m)
2
= 6.036E 24kg
6.036E 24
5.972E 24
= 1.0107Ear th Ma sses
R
m
= 2R
s
M
p
M
s
2(5.52E 8m)
(6.036E 24kg)
(1.5575E 30kg)
= 1.53656E6m =
1.53656E6m
1.7374E6m
= 0.8844Lu n arRa dii
of 48 53
=
Lets get the orbital velocity of this moon with equation 9:
So the orbital period of the moon is:
=
Compared to that of the Earth which is 27.3 days for the sidereal month and is a 29.53day lunar
month. The Moon of Tau Ceti would have a slightly longer lunar month, too. Now we can
determine the rotation period of the planet which would be its day:
We get the mass of the moon from equation 10:
=
So the rotation period of the planet (Its Day) is from equation 8:
=
=
If the moon of Tau Ceti is to perfectly eclipse its star, like our Moon does with the Sun, to let its
inhabitants know that they are there for reason, and so as to theoretically be a condition for
optimizing life, then we have the following should hold:
r
moon
= r
planet
R
moon
R
star
= (1.045E11m)
1.53656E6m
5.52E8m
= 2.91E8m
2.91E8m
3.84E8m
= 0.7578L u n ar OrbitalRa dii
v
m
=
GM
p
r
m
=
(6.67408E 11)(6.0636E 24kg
2.91E8m
= 1,176.6m /s
T
m
=
2π r
m
v
m
=
2π (.91E 8m)
1,176.6m /s
= 1.554E6secon d s
(1.554E6s)
min
60sec
h our
60min
d ay
24h ours
= 17.987d ays
M
m
=
s
τ
4r
2
m
=
(2.8314E 33J s)(0.5)(18429638s)
4(2.91E 8m)
2
= 7.7E 22kg
7.7E 22kg
7.34763E 22kg
= 1.048Ear th Moon s
Pl a n et Day =
2r
p
v
m
M
m
M
s
3
2(1.045E11m)
1,176.6m /s
7.7E 22kg
1.5575E 30kg
= 39495.61secon d s = 658.26min = 10.971h ours
10.971
24h ours
= 0.457Ear th Days
r
planet
r
moon
R
star
R
moon
of 49 53
We have
And it does hold. For the Earth/Moon/Sun system we get
For the Tau Ceti system the 359 is close to 360, which is a convenient amount of degrees into
which divide a circle, like we did. We see the Tau Cetians might do the same, because like us
they might choose the base 60 sexagesimal system of counting, and this 360 might influence the
way the inhabitants of a planet orbiting this star would make their calendar. Just like the 365
day year here on Earth corresponds closely to the 360 degrees of a circle. Aside from having here
on Earth the degree, we have gradians, of which there are 400 in a circle. They make it very easy
for computing right angles to a given angle. They are used more in Europe and in surveying and
architecture, though it might have some interesting connection with our 400=400.
The habitable planet of Tau Ceti would orbit it star with a period of 213 days, which would be its
year. It star is about 78% the mass of our Sun, and about 79% its size. Its habitable planet would
be a little larger than the Earth, but not much. Its Moon would be about 88% the size of our
moon, but a little bit more massive, but not much more massive. It would orbit Tau Ceti once
about every 18 days meaning there would be 11.833 months in the Tau Ceti Year, or about 12
months like we have. It would orbit the planet at about three quarters the distance ours does.
The day from sunrise to sunrise, or sunset to sunset would be about 11 hours, or close to half the
length of our day. This is if Tau Ceti was engineered for life, but it doesn’t have to have a planet
so optimally designed for life as ours. Our 24 hour day would be better for when entering the
realm of doing astronomy, because longer nights mean more complete observations of the
universe surrounding us than for the Tau Cetians. It may mean for the conditions for life to be
optimal, like here on Earth, the star would have to be as massive as the Sun, to provide a longer
year and a longer day. But we see here how the science of engineering planetary systems optimal
for life, might be done.
1.045E11m
2.91E8m
=
5.52E8m
1.53656E6m
359 = 359
400 = 400
of 50 53
Appendix 1
If we want to prove that our planetary Planck constant is correct, the delocalization time for the
Earth should be 6 months using it, the time for the Earth to travel the width of its orbit. We want
to solve the Schrödinger wave equation for a wave packet and use the most basic thing we can
which is a Gaussian distribution. We want to then substitute for Planck’s constant that is used
for quanta and atoms our Planck-type constant (h bar solar) for the Earth/Moon/Sun system
then apply it to predict the delocalization time for the Moon in its orbit with the Earth around
the Sun.
We consider a Gaussian wave-packet at t=0:
We say that is the delocalization length and decompose the wave packet with a Fourier
transform:
is the harmonics of the wave function. We use the identity that gives the integral of a
quadratic:
Solve the equation
With the initial condition
A plane wave is the solution:
Where,
The wave-packet evolves with time as
Calculate the Gaussian integral of
h
ψ (x,0) = Ae
x
2
2d
2
d
ψ (x,0) = Ae
x
2
2d
2
=
dp
2π
ϕ
p
e
i
px
ϕ
p
−∞
e
α
2
x+βx
d x =
π
α
e
β
2
4α
iℏ∂
t
ψ (x, t) =
p
2m
ψ (x, t)
ψ (x,0) =
dp e
p
2
d
2
2
2
e
i
px
e
i
( pxϵ( p)t)
ϵ( p) =
p
2
2m
ψ (x, t) =
dp e
p
2
d
2
2
2
e
i
( px
p
2
2m
t)
dp
of 51 53
and
The solution is:
where
is the delocalization distance, which for instance could be the width of an atom. is the
delocalization time, the average time for say an electron to traverse the diameter of the atom and
even leave it, to delocalize. If we substitute for our , and say that the delocalization distance
is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization
time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We
have
Where is the mass of the Moon, and is the orbital radius of the Moon. We have
Now let’s compute a half a year…
(1/2)(365.25)(24)(60)(60)=15778800 seconds
So we see our delocalization time is very close to the half year over which the Earth and Moon
travel from one position to the opposite side of the Sun. The closeness is
Thus we know our is accurate, it continues to function in a theoretical framework. The thing
about this is that it means we can predict the mass of the Moon from the Earth year. In terms of
what we said earlier that the Moon allows for life by creating the seasons, holding the Earth at
its tilt to the Sun, so we don’t go through extreme heat and cold, this suggests the Moon has a
mass that follows from Earth orbit, which is the habitable zone of the Sun, the right distance for
water to exist as liquid, and thus could be as it is for a reason, which means life might be part of
a physical process throughout the Universe, that it unfolds naturally in the evolution of star
systems.
α
2
=
d
2
2
2
+
it
2m
β =
i x
ψ
2
= ex p
[
x
2
d
2
1
1 + t
2
/τ
2
]
τ =
m d
2
d
τ
τ =
m
moon
(2r
moon
)
2
m
moon
r
moon
τ = 4
(7.34767E 22kg)(3.844E8m)
2
2.8314E 33J s
= 15338227secon d s
15338227
15778800
100 = 97.2 %
of 52 53
Appendix 2: The Data For Verifying The Equations
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Earth day=(24)(60)(60)=86,400 seconds. Using the Moons orbital velocity at aphelion, and
Earth’s orbital velocity at perihelion we have:
m
P
: 1.67262 × 10
27
kg
h : 6.62607 × 10
34
J s
r
p
: 0.833 × 10
15
m
G: 6.67408 × 10
11
N
m
2
kg
2
c : 299,792,458m /s
α : 1/137
q
p
= q
e
= 1.6022E 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 53 53
The Author!